Hence it follows that (a1 + a2 + … +an)/n≥ n√(a1a2 … an) with equality iff a1=a2=a3… = an. Thus P(n) is true and the result follows by the principle of Mathematical Induction.

Applications of the AGM to the sequence n1/n, where nεω, n≥1.

Numerical computation of the first few terms of this sequence shows that the first term 11/1 = 1, the second is 21/2 = √2 and the third term is 3√3 ( the maximum value of the sequence). For n≥3 the sequence appears to be monotone decreasing, tending to a limit of 1.

Firstly, 2 subsequent values cannot be equal. Otherwise n1/n = (n+1)1/(n+1) for some n. From this equation (if true) it follows that n(n+1) = (n+1)n. If n=1 then it follows that 1 = 2 which is obviously false. If n>1 then n and n+1 must have a common prime factor, again false since n and n+1 are coprime.

So, for all nεω, n≥1, either n1/n> (n+1)1/(n+1) or n1/n< (n+1)1/(n+1).